A math problem, the first year of junior high school 2 yuan 1 equation to solve, pay attention to th

by 5x-10y-4x-8y=-10

5x-4x-(10+8)y=-10

x-18y=-10

x=18y-10

Bring in Eq. 2.

1 3(18y-10)-2 3y-1 2(18y-10)-y=-1 to calculate the result of y. Count tired! Solve it yourself!

The method has taught you that the representative division sign is also the meaning of a fraction! The first day of this difficult mention, the blow.

5x-10y-4x-8y=-10

5x-4x-（10y+8y）=-10

x-18y=-10

I tried my best, that's all I can do, we haven't learned yet!!

5x-10y-4x-8y=-10

1÷3x-2÷3y-1÷2x-y=-1

Simplify. x-18y=-10①

y-4x-6xy²+6xy=0②

Result: x=18y-10

Substituting "yields: -y-4(18y-10)-6y(18y-10)+6y(18y-10)=0

Simplify: -133y+40-108y+168y=0I don't know if I did anything wrong, I calculated very seriously and hard for a long time, and actually simplified the cube and square!

I won't count the cubes... You don't have this kind of question in your first year of junior high school, do you? Did you make a mistake?

Solution: From (1), get.

x-18y=-10

From (2), derived.

4/3y-1/2x=-1

The highest order of unknowns in a 2-element 1st order equation is 1,This is not a 2-dimensional 1-order equation.

Hello! Solution: Let the length of one iron rod be x and the length of the other iron rod be y, then according to the question, {x+y=55,2 3x-4 5y=0

From the second equation, x=(6 5)y, substitute the first equation solution: x=30 y=25 So the water depth is (1-1 3)x=20cm Answer: The water depth in the bucket is 20cm. Good luck with your studies!

Solution: Let the lengths of the two iron rods be x and y respectively

x+y=55

x(1-1/3)=y(1-1/5)

Solution: x=30 y=25

x(1-1/3)=y(1-1/5)=20

A: At this time, the depth of water in the barrel is 20cm

Set: The depth of the water in the barrel is xcm

x divided by 2 3 + x divided by 4 5 = 55

x=20

The depth of the water in the barrel at this point is x centimeters.

x÷（1-1/5）+x÷（1-1/3）=555/4 x+3/2 x=55

11/4 x=55

x = 20 A: The depth of water in the barrel is 20 cm.

1.Draw a sketch. Let one of them be long x and the other long y. then x+y=55

2/3x=4/5y

Solve x,y, and you can get the depth.

2.Let the depth x, then.

x (2 3) + x (4 5) = 55 to get a depth of 20

Let the two sticks be long x and y respectively, by the title:

1-1/3）x=（1-1/5）y

x+y=55

The simultaneous solution is x=30 y=25

So the water depth is (1-1 3) x = 20cm

Let one rod be long x one long y, and the water depth is d

From the meaning of the title, x+y=55

2/3x=4/5y

Solution: x=30 y=25

Then the water depth d = 30 times 2 3 = 20cm

The lengths in the water are 2 3 and 4 5

So let's get that if the length of the two sticks is x,y, then x+y=55 (2 3)*x=(4 5)*y

It's good to solve it...

Set the depth of the water x cm, and the length of an iron rod is y cm

x=2/3y=4/5(55-y)

x=20

Solution: Assuming that the original planned time is x hours, then.

50（x+24/60）=75（x-24/60）50x+20=75x-30

25x=50

x = 2 The distance between A and B = 50 (2 + 24 60) = 120 (km).

Solution: Time is set to xThe distance is set to yThen there is 50(x+24)=y, and 75(x-24)=y, so we get: x=120, y=2

Set the distance of S kilometers and the specified time of T hours.

50(t+(24/60))=s

75(t-(24/60))=s

Solution: t=2 (hour) s=120 (km).

In fact, an equation can be used, if the specified time is x hours, then at a speed of 50 it takes x+ or 24 minutes) hours to arrive, and if it takes only an hour at a speed of 75.

50×(x+

If the system of equations is set to y, then y is the first day of the first year on either side of the above equation, which seems to be more difficult than I was at that time.

The main problem is that grasping the distance between two places is an invariant.

1. If each B-type machine produces x products per day, then each A-type machine produces x+1 products per day.

It is obtained from the problem and the equation.

5 (x+1)-4] 8 (7x-1) 11, i.e., 55x+11=56x-8

Solution, x 19

(7 19-1) 11 12 pcs per box.

So, 12 products per box.

2. A box of mooncakes needs flour

A total of 4,500 boxes of mooncakes can be produced.

Large moon cakes need flour

Small mooncakes need flour

Therefore, large moon cakes need 2500kg of flour, and small moon cakes need 2000kg of flour

1. Set x products in each box;

8x﹢4﹚÷5﹣﹙11x﹢1﹚÷7=1

7×﹙8x﹢4﹚﹣5×﹙11x﹢1﹚=3556x﹢28﹣55x﹣5=35

x﹢23=35

x=35﹣23

x=12

2. Set 4500kg of flour to make x boxes of moon cakes;

x=4500÷

x=25000 boxes

Flour is used to make large moon cakes2

Flour is used to make small mooncakes4

1) Set up a box of x pcs, then each type A machine has (8x+4) 5 products a day, and each type B machine has (11x+1) 7 products a day

Because each A-type machine produces 1 more product a day than the B-type machine, according to the problem, it is obtained that (8x+4) 5-(11x+1) 7=1, and the solution is x=12

2) Make xkg flour for large moon cakes, then (4500-x) kg flour for small moon cakes, according to the topic.

2 (x (2 times the number of large mooncakes = number of small mooncakes)

The solution is x=2500

4500-x=4500-2500=2000

(1) Set up a box of x pieces. Then (8x+4) 5-(11x+1) 7=1, the solution is x=12

Large moon cake 25000x2=50000

Small moon cakes 2000kg

Answer to the first question: The output of each type A machine is set to x, and the output of type B is set to y, and each box can hold x products.

a-b=15a=8x+4

7b=11x+1 x=12 a=20 b=19 Each box can hold 12 products.

Answer to the second question: If the flour required for the production of large boxes of flour is A, and the flour used for the production of small boxes of common flour is B, X boxes of flour can be produced.

a+b=4500

a=(2*b=(4* solution: a=2500 b=2000 x=25000

The large box needs 2500kg of flour, and the small box needs 2000kg of flour to produce a total of 25000 boxes of moon cakes.

The second question is because it is sold in boxes, so each box is required: 2 * Now there are 4500kg boxes (the maximum supporting output).

The large dosage is: 25000*2* flour.

The small dosage is: 25000*4* flour.

x+2y=9 (1)

3x-2y=-1 (2)

Obtained by calling Brother Li (1).

x-9=-2y （3）

Substitute (3) for (2) to get:

3x+x-9=-1

4x=8x=2 (4)

Substitute (4) for (1) to get:

x+2*2=9

x=5 so the solution of the original system of equations is.

x=5y=2

2)3u+2t=7 (1)

6u-2t=11 (2)

9u=18u=2 (3)

Substitute (3) with (1).

3*2+2t=7

t=1 2 and the solution of the original system of equations is.

u=2t=1/2

3)2a+b=3 (1)

3a+b=4 (2)

a=1 (3)

Substitute (3) with (1).

2*1+b=3

b=1, so the original scrambler solution is.

a=1b=1

(1) x+2y=9 ①

3x-2y=-1 ②

From the formula x=9-2y

Substituting the formula into the number of branches yields 3(9-2y)-2y=-1, i.e., 27-6y-2y=-1

8y=-28

y=7 2 substitute y=7 2 into the first of the celery to get x=2

So x=2

y=7／2(2) 3u+2t=7 ①

6u-2t=11 ②

9u=18 is obtained

u=2 substituting u=2 into , yielding 6+2t=7

t=1 2, so u=2

t=1／2(3) 2a+b=3 ①

3a+b=4 ②

The first dress gets a=1

Substituting a=1 into , we get 2+b=3

b=1, so a=1b=1

Solution: Substitution method: (1) X=9-2y can be known from x+2y=9, and x=9-2y is substituted into 3x-2y=-1 to obtain 3(9-2y)-2y=-1, then:

27-6y-2y=-1, move the unknown to the side of the equation, and get 8y=27+1, y=; Substituting y=into x=9-2y, we get x=2.

2) From 3U+2T=7, the shift term obtains 2T=7-3U, and substitutes 2T=7-3U into 6U-2T=11, then 6u-(7-3u)=11, then 6u-7+3u=11, so 9u-7=11, 9u=18, u=2;Substituting u=2 into 3u+2t=7, then 3*2+2t=7, 2t=1

3) Ibid. Outline.

Addition and subtraction: (1): Observation shows that there are 2y in both formulas, and the only reason for missing the observation is to use the elimination method to return to cultivation: add both sides of the equation x+2y=9 and 3x-2y=-1 at the same time to obtain: 4x=8, then x=2, substitute x=2 into any formula, and you can get y=

In the same way, the addition of the two formulas of (2) gives 9u=18, u=2, and substituting gives t=1;

3) is subtracted to give a=1, b=1

1.Let the unit price of MP4 be X yuan, and the unit price of the schoolbag will be Y yuan.

then x+y=452

x=4y-8

The solution is x=360 y=92

Therefore, the unit price of MP4 is 360 yuan, and the unit price of the schoolbag is 92 yuan.

2.You can buy it for 400 yuan at supermarket A.

After buying MP4 in B supermarket, you can get a 90 yuan shopping coupon, and you can also buy a schoolbag with 40 yuan in cash left.

But in supermarket A, you can have 400-452* yuan in cash left.

In supermarket B, you can leave 40 + 90-92 = 38 yuan.

So it's more cost-effective at supermarket A.

If my is helpful to you, please choose it as a satisfactory answer in time, thank you

Solution: (1) Set the unit price of MP4 to be X yuan, and the unit price of the schoolbag to be Y yuan.

System of equations: x+y=452

4y-8=x

Solution: x=360

y=92A: MP4** is 360 yuan, and the unit price of the schoolbag is 92 yuan.

Solution: (2) A: 452 yuan.

b: 400 100 = 4 4 30 = 120 yuan 120-(452-400) = 68 yuan.

400-68 = 332 yuan.

Mall B is more economical, and he can buy both goods.

One dollar at a time or two at a time? Binary equations are a system of equations, and this problem does not need a system of equations to calculate, since you have to ask for a binary one, let the speed of the plane be x and the wind speed be y. Time multiplied by speed and other distances, 5 2 hours multiplied by x = 1200, 10 3 (x + y) = 1200, solution, x = 480, y = -120

1. Analysis: According to the idea of engineering problems, the burning efficiency of the first candle is 1 60, and the burning efficiency of the second candle is 1 80

Because the first one burns quickly, if one of them is half of the other, it can only be that the first is half of the second.

After x minutes, we can give the equation: 1-x 60=(1-x 80) 2 gives x=48

The answer is 48 60=hours.

2. 1) If this team wins x games, it will draw (7-x) games.

3*x+1*（7-x）=17

x=52) has the highest score in the next 6 games, with a score of 17+3*6=35

The maximum score is 35 points.

3) Of all the situations where the goal is reached, the least number of games will be won except in the case of a tie. Set x number of wins.

then 3*x+1*(6-x)=29-17x=3